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Relativistic MomentumSummary:

Calculate relativistic momentum.

Explain why the only mass it makes sense to talk about is rest mass.

Figure 1: Momentum is an important concept forthese football players from the University ofCalifornia at Berkeley and the University ofCalifornia at Davis. Players with more mass oftenhave a larger impact because their momentum islarger. For objects moving at relativistic speeds,the effect is even greater. (credit: John MartinezPavliga)

In classical physics, momentum is a simple product of mass and velocity. However, we saw in thelast section that when special relativity is taken into account, massive objects have a speed limit.

What effect do you think mass and velocity have on the momentum of objects moving at relativistic

speeds?

Momentum is one of the most important concepts in physics. The broadest form of Newtons second

law is stated in terms of momentum. Momentum is conserved whenever the net external force on a

system is zero. This makes momentum conservation a fundamental tool for analyzing collisions. All

ofWork, Energy, and Energy Resourcesis devoted to momentum, and momentum has been

important for many other topics as well, particularly where collisions were involved. We will see that

momentum has the same importance in modern physics. Relativistic momentum is conserved, and

much of what we know about subatomic structure comes from the analysis of collisions of

accelerator-produced relativistic particles.
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The first postulate of relativity states that the laws of physics are the same in all inertial frames. Does

the law of conservation of momentum survive this requirement at high velocities? The answer is yes,

provided that the momentum is defined as follows.

RELATIVISTIC MOMENTUM:

Relativistic momentumpis classical momentum multiplied by the relativistic factor .

p=mu,

(1)

where mis the rest massof the object, uis its velocity relative to an observer, and the

relativistic factor

=11u2c2.

(2)

Note that we use ufor velocity here to distinguish it from relative velocity vbetween observers.

Only one observer is being considered here. Withpdefined in this way, total momentumptotis

conserved whenever the net external force is zero, just as in classical physics. Again we see that the

relativistic quantity becomes virtually the same as the classical at low velocities. That is, relativistic

momentum mubecomes the classical muat low velocities, because is very nearly equal to 1 at

low velocities.

Relativistic momentum has the same intuitive feel as classical momentum. It is greatest for large

masses moving at high velocities, but, because of the factor , relativistic momentum approaches

infinity as uapproaches c. (SeeFigure 2.) This is another indication that an object with mass cannotreach the speed of light. If it did, its momentum would become infinite, an unreasonable value.

Figure 2: Relativistic momentum approaches

infinity as the velocity of an object approaches

the speed of light.
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MISCONCEPTION ALERT: RELATIVISTIC MASS AND MOMENTUM:

The relativistically correct definition of momentum asp=muis sometimes taken to imply that

mass varies with velocity: mvar=m, particularly in older textbooks. However, note that mis themass of the object as measured by a person at rest relative to the object. Thus, mis defined to

be the rest mass, which could be measured at rest, perhaps using gravity. When a mass ismoving relative to an observer, the only way that its mass can be determined is through

collisions or other means in which momentum is involved. Since the mass of a moving object

cannot be determined independently of momentum, the only meaningful mass is rest mass.

Thus, when we use the term mass, assume it to be identical to rest mass.

Relativistic momentum is defined in such a way that the conservation of momentum will hold in all

inertial frames. Whenever the net external force on a system is zero, relativistic momentum is

conserved, just as is the case for classical momentum. This has been verified in numerous

experiments.

InRelativistic Energy,the relationship of relativistic momentum to energy is explored. That subject

will produce our first inkling that objects without mass may also have momentum.

Check Your Understanding

What is the momentum of an electron traveling at a speed 0.985c? The rest mass of the electron

is 9.111031kg.

[ Show ]

Section Summary The law of conservation of momentum is valid whenever the net external force is zero and

for relativistic momentum. Relativistic momentumpis classical momentum multiplied by the

relativistic factor .

p=mu, where mis the rest mass of the object, uis its velocity relative to an observer, and

the relativistic factor =11u2c2.

At low velocities, relativistic momentum is equivalent to classical momentum.

Relativistic momentum approaches infinity as uapproaches c. This implies that an object

with mass cannot reach the speed of light. Relativistic momentum is conserved, just as classical momentum is conserved.

Conceptual QuestionsExercise 1

How does modern relativity modify the law of conservation of momentum?
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Exercise 2

Is it possible for an external force to be acting on a system and relativistic momentum to be

conserved? Explain.

Problem ExercisesExercise 1Find the momentum of a helium nucleus having a mass of 6.681027kgthat is moving at 0.200c.

Solut ion

4.091019kgm/s

Exercise 2

What is the momentum of an electron traveling at 0.980c?

Exercise 3

(a) Find the momentum of a 1.00109kgasteroid heading towards the Earth at 30.0 km/s. (b)

Find the ratio of this momentum to the classical momentum. (Hint: Use the approximation

that =1+(1/2)v2/c2at low velocities.)

Solut ion

(a) 3.0000000151013kgm/s.

(b) Ratio of relativistic to classical momenta equals 1.000000005 (extra digits to show small effects)

Exercise 4

(a) What is the momentum of a 2000 kg satellite orbiting at 4.00 km/s? (b) Find the ratio of this

momentum to the classical momentum. (Hint: Use the approximation that =1+(1/2)v2/c2atlow velocities.)Exercise 5

What is the velocity of an electron that has a momentum of 3.041021kgm/s? Note that you must

calculate the velocity to at least four digits to see the difference from c.

Solut ion

2.9957108m/s

Exercise 6

Find the velocity of a proton that has a momentum of 4.481019kgm/s.


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Exercise 7

(a) Calculate the speed of a 1.00-gparticle of dust that has the same momentum as a proton

moving at 0.999c. (b) What does the small speed tell us about the mass of a proton compared to

even a tiny amount of macroscopic matter?

Solut ion

(a) 1.121108m/s

(b) The small speed tells us that the mass of a proton is substantially smaller than that of even a tiny

amount of macroscopic matter!

Solut ion

(a) 1.121108m/s

(b) The small speed tells us that the mass of a proton is substantially smaller than that of even a tiny

amount of macroscopic matter!

Exercise 8

(a) Calculate for a proton that has a momentum of 1.00 kgm/s.(b) What is its speed? Such

protons form a rare component of cosmic radiation with uncertain origins.

Glossary

relativistic momentum:

p, the momentum of an object moving at relativistic velocity;p=mu, where mis the rest

mass of the object, uis its velocity relative to an observer, and the relativistic

factor =11u2c2

rest mass:

the mass of an object as measured by a person at rest relative to the object

Relativistic Energy

Module by:OpenStax College.E-mail the author

Summary:
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Compute total energy of a relativistic object.

Compute the kinetic energy of a relativistic object.

Describe rest energy, and explain how it can be converted to other forms.

Explain why massive particles cannot reach C.

Figure 1: The National Spherical Torus

Experiment (NSTX) has a fusion reactor in

which hydrogen isotopes undergo fusion to

produce helium. In this process, a relatively

small mass of fuel is converted into a large

amount of energy. (credit: Princeton Plasma

Physics Laboratory)

A tokamak is a form of experimental fusion reactor, which can change mass

to energy. Accomplishing this requires an understanding of relativistic

energy. Nuclear reactors are proof of the conservation of relativistic energy.

Conservation of energy is one of the most important laws in physics. Not

only does energy have many important forms, but each form can be

converted to any other. We know that classically the total amount of energy

in a system remains constant. Relativistically, energy is still conserved,

provided its definition is altered to include the possibility of mass changing toenergy, as in the reactions that occur within a nuclear reactor. Relativistic

energy is intentionally defined so that it will be conserved in all inertial

frames, just as is the case for relativistic momentum. As a consequence, we

learn that several fundamental quantities are related in ways not known in

classical physics. All of these relationships are verified by experiment and


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have fundamental consequences. The altered definition of energy contains

some of the most fundamental and spectacular new insights into nature found

in recent history.

Total Energy and Rest EnergyThe first postulate of relativity states that the laws of physics are the same in

all inertial frames. Einstein showed that the law of conservation of energy is

valid relativistically, if we define energy to include a relativistic factor.

TOTAL ENERGY:

Total energyEis defined to be

E=mc2,

(1)

where mis mass, cis the speed of light, =11v2c2, and vis the velocity of the

mass relative to an observer. There are many aspects of the total

energyEthat we will discussamong them are how kinetic and potential

energies are included inE, and howEis related to relativistic momentum. But

first, note that at rest, total energy is not zero. Rather, when v=0, we

have =1, and an object has rest energy.

REST ENERGY:

Rest energyis

E0=mc2.

(2)

This is the correct form of Einsteins most famous equation, which for thefirst time showed that energy is related to the mass of an object at rest. For

example, if energy is stored in the object, its rest mass increases. This alsoimplies that mass can be destroyed to release energy. The implications of

these first two equations regarding relativistic energy are so broad that they

were not completely recognized for some years after Einstein published them

in 1907, nor was the experimental proof that they are correct widely


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recognized at first. Einstein, it should be noted, did understand and describe

the meanings and implications of his theory.

Example 1: Calculating Rest Energy: Rest Energy is Very Large

Calculate the rest energy of a 1.00-g mass.

Strategy

One gram is a small massless than half the mass of a penny. We canmultiply this mass, in SI units, by the speed of light squared to find the

equivalent rest energy.

Solution

1. Identify the knowns. m=1.00103kg; c=3.00108m/s

2. Identify the unknown.E03. Choose the appropriate equation.E0=mc2

4. Plug the knowns into the equation.

E0==mc2=(1.00103kg)(3.00108m/s)29.001013kgm2/s2

(3)

5. Convert units.

Noting that 1kgm2/s2=1 J, we see the rest mass energy is

E0=9.001013J.

(4)

Discussion

This is an enormous amount of energy for a 1.00-g mass. We do not notice

this energy, because it is generally not available. Rest energy is large because

the speed of light cis a large number and c2is a very large number, so

that mc2is huge for any macroscopic mass. The 9.001013Jrest mass energy

for 1.00 g is about twice the energy released by the Hiroshima atomic bomband about 10,000 times the kinetic energy of a large aircraft carrier. If a way

can be found to convert rest mass energy into some other form (and all forms

of energy can be converted into one another), then huge amounts of energy

can be obtained from the destruction of mass.


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Today, the practical applications of the conversion of mass into another form

of energy, such as in nuclear weapons and nuclear power plants, are well

known. But examples also existed when Einstein first proposed the correct

form of relativistic energy, and he did describe some of them. Nuclear

radiation had been discovered in the previous decade, and it had been amystery as to where its energy originated. The explanation was that, in

certain nuclear processes, a small amount of mass is destroyed and energy is

released and carried by nuclear radiation. But the amount of mass destroyed

is so small that it is difficult to detect that any is missing. Although Einstein

proposed this as the source of energy in the radioactive salts then being

studied, it was many years before there was broad recognition that mass

could be and, in fact, commonly is converted to energy. (SeeFigure 2.)

Figure 2: The Sun (a) and the Susquehanna

Steam Electric Station (b) both convert mass

into energythe Sun via nuclear fusion, the

electric station via nuclear fission. (credits: (a)

NASA/Goddard Space Flight Center, Scientific

Visualization Studio; (b) U.S. government)
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Because of the relationship of rest energy to mass, we now consider mass to

be a form of energy rather than something separate. There had not even been

a hint of this prior to Einsteins work. Such conversion is now known to bethe source of the Suns energy, the energy of nuclear decay, and even the

source of energy keeping Earths interior hot.

Stored Energy and Potential Energy

What happens to energy stored in an object at rest, such as the energy put into

a battery by charging it, or the energy stored in a toy guns compressedspring? The energy input becomes part of the total energy of the object and,

thus, increases its rest mass. All stored and potential energy becomes mass in

a system. Why is it we dont ordinarily notice this? In fact, conservation ofmass (meaning total mass is constant) was one of the great laws verified by

19th-century science. Why was it not noticed to be incorrect? The following

example helps answer these questions.

Example 2: Calculating Rest Mass: A Small Mass Increase due to Energy Input

A car battery is rated to be able to move 600 ampere-hours (Ah)of charge at

12.0 V. (a) Calculate the increase in rest mass of such a battery when it is

taken from being fully depleted to being fully charged. (b) What percent

increase is this, given the batterys mass is 20.0 kg?Strategy

In part (a), we first must find the energy stored in the battery, which equals

what the battery can supply in the form of electrical potential energy.

Since PEelec=qV, we have to calculate the charge qin 600Ah, which is the

product of the currentIand the time t. We then multiply the result by 12.0 V.

We can then calculate the batterys increase in mass using E=PEelec=(m)c2.Part (b) is a simple ratio converted to a percentage.

Solution for (a)

1. Identify the knowns.It=600 Ah; V=12.0V; c=3.00108m/s

2. Identify the unknown. m3. Choose the appropriate equation. PEelec=(m)c24. Rearrange the equation to solve for the unknown. m=PEelecc2


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m====PEelecc2qVc2(It)Vc2(600 Ah)(12.0 V)(3.00108)2.(5)

Write amperes A as coulombs per second (C/s), and convert hours toseconds.

m==(600 C/sh(3600 s1 h)(12.0 J/C)(3.00108m/s)2(2.16106C)(12.0J/C)(3.00108m/s)2

(6)

Using the conversion 1kgm2/s2=1J, we can write the mass as

m=2.881010kg.

Solution for (b)

1. Identify the knowns. m=2.881010kg; m=20.0 kg2. Identify the unknown. % change

3. Choose the appropriate equation. % increase=mm100%


% increase===mm100%2.881010kg20.0 kg100%1.44109%.(7)

Discussion

Both the actual increase in mass and the percent increase are very small,

since energy is divided by c2, a very large number. We would have to be able

to measure the mass of the battery to a precision of a billionth of a percent, or

1 part in 1011, to notice this increase. It is no wonder that the mass variation is

not readily observed. In fact, this change in mass is so small that we may

question how you could verify it is real. The answer is found in nuclear

processes in which the percentage of mass destroyed is large enough to bemeasured. The mass of the fuel of a nuclear reactor, for example, is

measurably smaller when its energy has been used. In that case, stored energy

has been released (converted mostly to heat and electricity) and the rest mass

has decreased. This is also the case when you use the energy stored in a


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battery, except that the stored energy is much greater in nuclear processes,

making the change in mass measurable in practice as well as in theory.

Kinetic Energy and the Ultimate Speed

LimitKinetic energy is energy of motion. Classically, kinetic energy has the

familiar expression 12mv2. The relativistic expression for kinetic energy is

obtained from the work-energy theorem. This theorem states that the net

work on a system goes into kinetic energy. If our system starts from rest, then

the work-energy theorem is

Wnet=KE.

(8)

Relativistically, at rest we have rest energyE0=mc2. The work increases this

to the total energyE=mc2. Thus,

Wnet=EE0=mc2mc2=(1)mc2.

(9)

Relativistically, we have Wnet=KErel.

RELATIVISTIC KINETIC ENERGY:

Relativistic kinetic energyis

KErel=(1)mc2.

(10)

When motionless, we have v=0and

=11v2c2=1,

(11)

so that KErel=0at rest, as expected. But the expression for relativistic kinetic

energy (such as total energy and rest energy) does not look much like the

classical 12mv2. To show that the classical expression for kinetic energy is

obtained at low velocities, we note that the binomial expansion for at low

velocities gives


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=1+12v2c2.

(12)

A binomial expansion is a way of expressing an algebraic quantity as a sum

of an infinite series of terms. In some cases, as in the limit of small velocityhere, most terms are very small. Thus the expression derived for here is not

exact, but it is a very accurate approximation. Thus, at low velocities,

1=12v2c2.

(13)

Entering this into the expression for relativistic kinetic energy gives

KErel

=[12v2

c2

]mc2

=12mv2

=KEclass

.

(14)

So, in fact, relativistic kinetic energy does become the same as classical

kinetic energy when vmc2. (b)IsEpcwhen =30.0, as for the astronaut discussed in the twin paradox?Solut ion

(a)E2=p2c2+m2c4=2m2c4,so thatp2c2=(21)m2c4, andtherefore(pc)2(mc2)2=21

(b) yes

[ Hide Solution ]
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Exercise 18

One cosmic ray neutron has a velocity of 0.250crelative to the Earth. (a) What is the neutrons total

energy in MeV? (b) Find its momentum. (c) IsEpcin this situation? Discuss in terms of theequation given in part (a) of the previous problem.

Exercise 19What is for a proton having a mass energy of 938.3 MeV accelerated through an effective potential

of 1.0 TV (teravolt) at Fermilab outside Chicago?

Solut ion

1.07103[ Hide Solution ]

Exercise 20

(a) What is the effective accelerating potential for electrons at the Stanford Linear Accelerator,

if =1.00105for them? (b) What is their total energy (nearly the same as kinetic in this case) in

GeV?

Exercise 21

(a) Using data from(Reference),find the mass destroyed when the energy in a barrel of crude oil is

released. (b) Given these barrels contain 200 liters and assuming the density of crude oil is 750

kg/m3, what is the ratio of mass destroyed to original mass, m/m?Solut ion

6.56108kg

4.371010

[ Hide Solution ]

Exercise 22

(a) Calculate the energy released by the destruction of 1.00 kg of mass. (b) How many kilograms

could be lifted to a 10.0 km height by this amount of energy?

Exercise 23

A Van de Graaff accelerator utilizes a 50.0 MV potential difference to accelerate charged particles

such as protons. (a) What is the velocity of a proton accelerated by such a potential? (b) An electron?

Solut ion

0.314c

0.99995c[ Hide Solution ]
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Exercise 24

Suppose you use an average of 500 kWhof electric energy per month in your home. (a) How long

would 1.00 g of mass converted to electric energy with an efficiency of 38.0% last you? (b) How

many homes could be supplied at the 500 kWhper month rate for one year by the energy from the

described mass conversion?Exercise 25

(a) A nuclear power plant converts energy from nuclear fission into electricity with an efficiency of

35.0%. How much mass is destroyed in one year to produce a continuous 1000 MW of electric

power? (b) Do you think it would be possible to observe this mass loss if the total mass of the fuel

is 104kg?

Solut ion

(a) 1.00 kg

(b) This much mass would be measurable, but probably not observable just by looking because it is

0.01% of the total mass.

[ Hide Solution ]

Exercise 26

Nuclear-powered rockets were researched for some years before safety concerns became paramount.

(a) What fraction of a rockets mass would have to be destroyed to get it into a low Earth orbit,

neglecting the decrease in gravity? (Assume an orbital altitude of 250 km, and calculate both the

kinetic energy (classical) and the gravitational potential energy needed.) (b) If the ship has a mass

of 1.00105kg(100 tons), what total yield nuclear explosion in tons of TNT is needed?

Exercise 27

The Sun produces energy at a rate of 4.001026W by the fusion of hydrogen. (a) How many

kilograms of hydrogen undergo fusion each second? (b) If the Sun is 90.0% hydrogen and half of this

can undergo fusion before the Sun changes character, how long could it produce energy at its current

rate? (c) How many kilograms of mass is the Sun losing per second? (d) What fraction of its mass

will it have lost in the time found in part (b)?

Solut ion

(a) 6.31011kg/s

(b) 4.51010y(c) 4.44109kg

(d) 0.32%
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